3.280 \(\int e^{\coth ^{-1}(x)} (1+x) \, dx\)

Optimal. Leaf size=79 \[ \frac{1}{2} \left (\frac{1}{x}+1\right )^{3/2} \sqrt{\frac{x-1}{x}} x^2+\frac{3}{2} \sqrt{\frac{1}{x}+1} \sqrt{\frac{x-1}{x}} x+\frac{3}{2} \tanh ^{-1}\left (\sqrt{\frac{1}{x}+1} \sqrt{\frac{x-1}{x}}\right ) \]

[Out]

(3*Sqrt[1 + x^(-1)]*Sqrt[(-1 + x)/x]*x)/2 + ((1 + x^(-1))^(3/2)*Sqrt[(-1 + x)/x]*x^2)/2 + (3*ArcTanh[Sqrt[1 +
x^(-1)]*Sqrt[(-1 + x)/x]])/2

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Rubi [A]  time = 0.046782, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {6175, 6180, 94, 92, 206} \[ \frac{1}{2} \left (\frac{1}{x}+1\right )^{3/2} \sqrt{\frac{x-1}{x}} x^2+\frac{3}{2} \sqrt{\frac{1}{x}+1} \sqrt{\frac{x-1}{x}} x+\frac{3}{2} \tanh ^{-1}\left (\sqrt{\frac{1}{x}+1} \sqrt{\frac{x-1}{x}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCoth[x]*(1 + x),x]

[Out]

(3*Sqrt[1 + x^(-1)]*Sqrt[(-1 + x)/x]*x)/2 + ((1 + x^(-1))^(3/2)*Sqrt[(-1 + x)/x]*x^2)/2 + (3*ArcTanh[Sqrt[1 +
x^(-1)]*Sqrt[(-1 + x)/x]])/2

Rule 6175

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> Dist[d^p, Int[u*x^p*(1 + c/(d*
x))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[a^2*c^2 - d^2, 0] &&  !IntegerQ[n/2] && Inte
gerQ[p]

Rule 6180

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_.), x_Symbol] :> -Dist[c^p, Subst[Int[((1
+ (d*x)/c)^p*(1 + x/a)^(n/2))/(x^(m + 2)*(1 - x/a)^(n/2)), x], x, 1/x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ
[c^2 - a^2*d^2, 0] &&  !IntegerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) && IntegerQ[m]

Rule 94

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

Rule 92

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))), x_Symbol] :> Dist[b*f, Subst[I
nt[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sqrt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
EqQ[2*b*d*e - f*(b*c + a*d), 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int e^{\coth ^{-1}(x)} (1+x) \, dx &=\int e^{\coth ^{-1}(x)} \left (1+\frac{1}{x}\right ) x \, dx\\ &=-\operatorname{Subst}\left (\int \frac{(1+x)^{3/2}}{\sqrt{1-x} x^3} \, dx,x,\frac{1}{x}\right )\\ &=\frac{1}{2} \left (1+\frac{1}{x}\right )^{3/2} \sqrt{\frac{-1+x}{x}} x^2-\frac{3}{2} \operatorname{Subst}\left (\int \frac{\sqrt{1+x}}{\sqrt{1-x} x^2} \, dx,x,\frac{1}{x}\right )\\ &=\frac{3}{2} \sqrt{1+\frac{1}{x}} \sqrt{-\frac{1-x}{x}} x+\frac{1}{2} \left (1+\frac{1}{x}\right )^{3/2} \sqrt{\frac{-1+x}{x}} x^2-\frac{3}{2} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} x \sqrt{1+x}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{3}{2} \sqrt{1+\frac{1}{x}} \sqrt{-\frac{1-x}{x}} x+\frac{1}{2} \left (1+\frac{1}{x}\right )^{3/2} \sqrt{\frac{-1+x}{x}} x^2+\frac{3}{2} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt{1+\frac{1}{x}} \sqrt{\frac{-1+x}{x}}\right )\\ &=\frac{3}{2} \sqrt{1+\frac{1}{x}} \sqrt{-\frac{1-x}{x}} x+\frac{1}{2} \left (1+\frac{1}{x}\right )^{3/2} \sqrt{\frac{-1+x}{x}} x^2+\frac{3}{2} \tanh ^{-1}\left (\sqrt{1+\frac{1}{x}} \sqrt{-\frac{1-x}{x}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0240794, size = 40, normalized size = 0.51 \[ \frac{1}{2} \sqrt{1-\frac{1}{x^2}} x (x+4)+\frac{3}{2} \log \left (\left (\sqrt{1-\frac{1}{x^2}}+1\right ) x\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcCoth[x]*(1 + x),x]

[Out]

(Sqrt[1 - x^(-2)]*x*(4 + x))/2 + (3*Log[(1 + Sqrt[1 - x^(-2)])*x])/2

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Maple [A]  time = 0.104, size = 57, normalized size = 0.7 \begin{align*}{\frac{-1+x}{2} \left ( x\sqrt{{x}^{2}-1}+4\,\sqrt{{x}^{2}-1}+3\,\ln \left ( x+\sqrt{{x}^{2}-1} \right ) \right ){\frac{1}{\sqrt{{\frac{-1+x}{1+x}}}}}{\frac{1}{\sqrt{ \left ( 1+x \right ) \left ( -1+x \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((-1+x)/(1+x))^(1/2)*(1+x),x)

[Out]

1/2*(-1+x)*(x*(x^2-1)^(1/2)+4*(x^2-1)^(1/2)+3*ln(x+(x^2-1)^(1/2)))/((-1+x)/(1+x))^(1/2)/((1+x)*(-1+x))^(1/2)

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Maxima [A]  time = 1.04258, size = 117, normalized size = 1.48 \begin{align*} \frac{3 \, \left (\frac{x - 1}{x + 1}\right )^{\frac{3}{2}} - 5 \, \sqrt{\frac{x - 1}{x + 1}}}{\frac{2 \,{\left (x - 1\right )}}{x + 1} - \frac{{\left (x - 1\right )}^{2}}{{\left (x + 1\right )}^{2}} - 1} + \frac{3}{2} \, \log \left (\sqrt{\frac{x - 1}{x + 1}} + 1\right ) - \frac{3}{2} \, \log \left (\sqrt{\frac{x - 1}{x + 1}} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((-1+x)/(1+x))^(1/2)*(1+x),x, algorithm="maxima")

[Out]

(3*((x - 1)/(x + 1))^(3/2) - 5*sqrt((x - 1)/(x + 1)))/(2*(x - 1)/(x + 1) - (x - 1)^2/(x + 1)^2 - 1) + 3/2*log(
sqrt((x - 1)/(x + 1)) + 1) - 3/2*log(sqrt((x - 1)/(x + 1)) - 1)

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Fricas [A]  time = 1.91277, size = 158, normalized size = 2. \begin{align*} \frac{1}{2} \,{\left (x^{2} + 5 \, x + 4\right )} \sqrt{\frac{x - 1}{x + 1}} + \frac{3}{2} \, \log \left (\sqrt{\frac{x - 1}{x + 1}} + 1\right ) - \frac{3}{2} \, \log \left (\sqrt{\frac{x - 1}{x + 1}} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((-1+x)/(1+x))^(1/2)*(1+x),x, algorithm="fricas")

[Out]

1/2*(x^2 + 5*x + 4)*sqrt((x - 1)/(x + 1)) + 3/2*log(sqrt((x - 1)/(x + 1)) + 1) - 3/2*log(sqrt((x - 1)/(x + 1))
- 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x + 1}{\sqrt{\frac{x - 1}{x + 1}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((-1+x)/(1+x))**(1/2)*(1+x),x)

[Out]

Integral((x + 1)/sqrt((x - 1)/(x + 1)), x)

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Giac [A]  time = 1.11125, size = 113, normalized size = 1.43 \begin{align*} -\frac{\frac{3 \,{\left (x - 1\right )} \sqrt{\frac{x - 1}{x + 1}}}{x + 1} - 5 \, \sqrt{\frac{x - 1}{x + 1}}}{{\left (\frac{x - 1}{x + 1} - 1\right )}^{2}} + \frac{3}{2} \, \log \left (\sqrt{\frac{x - 1}{x + 1}} + 1\right ) - \frac{3}{2} \, \log \left ({\left | \sqrt{\frac{x - 1}{x + 1}} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((-1+x)/(1+x))^(1/2)*(1+x),x, algorithm="giac")

[Out]

-(3*(x - 1)*sqrt((x - 1)/(x + 1))/(x + 1) - 5*sqrt((x - 1)/(x + 1)))/((x - 1)/(x + 1) - 1)^2 + 3/2*log(sqrt((x
- 1)/(x + 1)) + 1) - 3/2*log(abs(sqrt((x - 1)/(x + 1)) - 1))